Assuming 100% dissociation of 1 m KCl soluti | vedclass

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(B) The freezing point depression is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since $KCl$ dissociates completely as $KCl ightarrow

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$-3.72$

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(B) The freezing point depression is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since $KCl$ dissociates completely as $KCl ightarrow K^+ + Cl^-$,the van't Hoff factor $i = 2$. Given molality $m = 1 \ m$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$. $\Delta T_f = 2 	imes 1.86 	imes 1 = 3.72 \ K$. The freezing point of pure water is $0 \ ^oC$ $(273.15 \ K)$. Freezing point of solution = $T_f^o - \Delta T_f = 0 \ ^oC - 3.72 \ ^oC = -3.72 \ ^oC$.

Assuming $100\%$ dissociation of $1 \ m$ $KCl$ solution in water,the freezing point of the solution will be ........ $^oC$. $(K_f = 1.86 \ K \ kg \ mol^{-1})$

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